Question

Can someone correct me? I can't figure out what I'm doing wrong... (posting problem below)

Answer 1

I have to find the critical numbers for the function f(x) = ((x+1)^3)(x-3)
I took the derivative and I got


f'(x) = (x+1)(3x-9)(x+1)^2


which would mean my critical numbers are x = -1 and x -3

But the graph says that there should only be one relative extreme ( a relative min at (2,-27))

Am i missing something?

Answer 2

I meant that the critical numbers are at x = -1 and x = 3*

Answer 3

Did you find this derivative using the product rule?

Answer 4

Yup

Answer 5

I did f'(x) = (x+1)^3 * 1 + (x-3) * 3(x+1)^2

Answer 6

\[ (x+1)^3 * 1 + (x-3) * 3(x+1)^2 = 0 \\
(x+1)^2 ( x+1 +3(x-3)) = 0 \\
(x+1)^2 (4x-8) = 0
\]

Answer 7

How did you get from from the first line to the second line?

Answer 8

Ohhh wait I forgot about that rule. I multiplied (x+1) and (3x-9) when I should have added them

Answer 9

x=-1 is a critical point, but it's not a min or max, it's an inflection point

*factor (x+1)^2 out from each term

Answer 10

I get it now, thanks a lot

Answer 11

It's always good to eyeball a graph (I used geogebra to plot. It's free)